A) \[\text{100}\sqrt{3}\]
B) \[50\sqrt{2}\]
C) 100
D) 50
Correct Answer: C
Solution :
Let AB=h, \[\text{QR=2a}\] \[\text{In }\Delta \text{ABQ, tan30}{}^\circ \text{=}\frac{AB}{QB}=\frac{h}{a}\Rightarrow a=h\sqrt{3}\] \[\text{In }\Delta \text{PBA, tan45}{}^\circ \text{=}\frac{AB}{PB}=\frac{h}{\sqrt{{{200}^{2}}-{{a}^{2}}}}\] \[\Rightarrow 1=\frac{h}{\sqrt{{{200}^{2}}-3{{h}^{2}}}}\] \[\Rightarrow 4{{h}^{2}}={{200}^{2}}\] \[\Rightarrow h=100\]You need to login to perform this action.
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